Question

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 80° − sec 10° = 0
(iv) cosec²72° − tan²18° = 1
(v) cos²75° + cos²15° = 1
(vi) tan²66° − cot²24° = 0
(vii) sin²48° + sin²42° = 1
(viii) cos²57° − sin²33° = 0
(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \left(\text{i}\right){\text{LHS=cos81}}^0-\sin 9^0\\ \end{array} \\ \begin{array}{l}\text{=cos}({90}^0-9^0)-\sin 9^0\\ \end{array} \\ \begin{array}{l}{\text{=sin9}}^0-\sin 9^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{ii}\right){\text{LHS=tan71}}^0-\cot {19}^0\\ \end{array} \\ \begin{array}{l}\text{=tan}({90}^0-{19}^0)-\cot {19}^0\\ \end{array} \\ \begin{array}{l}{\text{=cot19}}^0-\cot {19}^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \begin{array}{l}\left(\text{iii}\right){\text{LHS=cosec80}}^0-\sec {10}^0\\ \end{array}\end{array} \\ \begin{array}{l}\text{=cosec}({90}^0-{10}^0)-\sec {10}^0\\ \end{array} \\ \begin{array}{l}{\text{=sec10}}^0-\sec {10}^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{iv}\right){\text{LHS=cosec}}^2{72}^0-{\tan }^2{18}^0\\ \end{array} \\ \begin{array}{l}{\text{=cosec}}^2({90}^0-{18}^0)-{\tan }^2{18}^0\\ \end{array} \\ \begin{array}{l}{\text{=sec}}^2{18}^0-{\tan }^2{18}^0\\ \end{array} \\ \begin{array}{l}\text{=1}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{v}\right){\text{LHS=cos}}^2{75}^0+{\cos }^2{15}^0\\ \end{array} \\ \begin{array}{l}{\text{=cos}}^2({90}^0-{15}^0)+{\cos }^2{15}^0\\ \end{array} \\ \begin{array}{l}{\text{=sin}}^2{15}^0+{\cos }^2{15}^0\\ \end{array} \\ \text{=1}\text{.} \\ \text{=RHS} \\ \begin{array}{l}\\ \left(\text{vi}\right){\text{LHS=tan}}^2{66}^0-{\cot }^2{24}^0\\ \end{array} \\ \begin{array}{l}{\text{=tan}}^2({90}^0-{24}^0)-{\cot }^2{24}^0\\ \end{array} \\ \begin{array}{l}{\text{=cot}}^2{24}^0-{\cot }^2{24}^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{vii}\right){\text{LHS=sin}}^2{48}^0+{\sin }^2{42}^0\\ \end{array} \\ \begin{array}{l}{\text{=sin}}^2({90}^0-{42}^0)+{\sin }^2{42}^0\\ \end{array} \\ \begin{array}{l}{\text{=cos}}^2{42}^0+{\sin }^2{42}^0\\ \end{array} \\ \begin{array}{l}\text{=1}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{viii}\right){\text{LHS=cos}}^2{57}^0-{\sin }^2{33}^0\\ \end{array} \\ \begin{array}{l}{\text{=cos}}^2({90}^0-{33}^0)-{\sin }^2{33}^0\\ \end{array} \\ \begin{array}{l}{\text{=sin}}^2{33}^0-{\sin }^2{33}^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \begin{array}{l}\text{=RHS}\\ \end{array} \\ \begin{array}{l}\left(\text{ix}\right)\text{LHS=}({\text{sin65}}^0+\cos {25}^0)({\text{sin65}}^0-\cos {25}^0)\\ \end{array} \\ \begin{array}{l}{\text{=sin}}^2{65}^0-{\cos }^2{25}^0\\ \end{array} \\ \begin{array}{l}{\text{=sin}}^2({90}^0-{25}^0)-{\cos }^2{25}^0\\ \end{array} \\ \begin{array}{l}{\text{=cos}}^2{25}^0-\cos {25}^0\\ \end{array} \\ \begin{array}{l}\text{=0}\\ \end{array} \\ \text{=RHS} \end{gathered}$$